Here is the question im stuck on.

solve sec^2 x - (1+rt3)tan x + rt3 = 1 for the range 0< x < 4pi giving the answer in terms of pi.

Thanks

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- Oct 1st 2007, 12:27 PM #1

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- Oct 1st 2007, 06:37 PM #2

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$\displaystyle \sec^2 x = \tan^2x +1$

Thus,

$\displaystyle \tan^2 x + 1 - (1+\sqrt{3})\tan x + \sqrt{3} = 1$

Thus,

$\displaystyle \tan^2 x -(1+\sqrt{3})\tan x + \sqrt{3} = 0$

Let $\displaystyle y=\tan x$ to get,

$\displaystyle y^2 - (1+\sqrt{3})y+\sqrt{y}=0$.

Use quadradic forumula.